# ques no 18 with explaination

$U=\frac{A\sqrt{x}}{{x}^{2}+B}...1\phantom{\rule{0ex}{0ex}}1)As{x}^{2}isaddedtoBsoasliketermshasthesamedimension.\phantom{\rule{0ex}{0ex}}dimensionofB={L}^{2}\phantom{\rule{0ex}{0ex}}2)Dimensionofdenomenatoris={L}^{2}\phantom{\rule{0ex}{0ex}}U=\frac{A\sqrt{x}}{{x}^{2}+B}\phantom{\rule{0ex}{0ex}}U=\frac{A\sqrt{x}}{{L}^{2}}\phantom{\rule{0ex}{0ex}}A=\frac{U\times {L}^{2}}{\sqrt{x}}\phantom{\rule{0ex}{0ex}}A=\frac{M{L}^{2}{T}^{-2}\times {L}^{2}}{\sqrt{L}}\phantom{\rule{0ex}{0ex}}A=M{L}^{(4-\frac{1}{2})}{T}^{-2}\phantom{\rule{0ex}{0ex}}A=M{L}^{\frac{7}{2}}{T}^{-2}\phantom{\rule{0ex}{0ex}}DimensionofAB=M{L}^{\frac{7}{2}}{T}^{-2}\times {L}^{2}\phantom{\rule{0ex}{0ex}}DimensionofAB=M{L}^{(\frac{7}{2}+2)}{T}^{-2}\phantom{\rule{0ex}{0ex}}DimensionofAB=M{L}^{\frac{11}{2}}{T}^{-2}\phantom{\rule{0ex}{0ex}}Regards\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

**
**